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It is my understanding that the Abel-Ruffini Theorem implies that certain polynomial equations $(x^5-x+1=0$, for instance) have transcendental roots. However, the Fundamental Theorem of Algebra states that we can factorise any polynomial into quadratics, and we can then solve these with the quadratic equation (implying that the solution would be algebraic).

On second thought, the only way I can see this to be resolved is if the quadratic factors of our polynomials have transcendental coefficients. Is this the case?

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The roots of $x^5-x+1=0$ are by definition algebraic. –  André Nicolas Aug 28 '12 at 15:32
    
I think youve misunderstood quite a few things: the A-R theorem doesn't state what you say, which is an oxymoron since a transcendental element, per definition, can't be root of of rational polynomials. Second, the FTA doesn't state that any (rational or real) polynomial can be factored in quadratic, which is false (on the rationals), what it says is that teh complex field $\,\Bbb C\,$ is algebraically closed, i.e.: every complex polynomial factors, over the complex, in linear factors. –  DonAntonio Aug 28 '12 at 15:34
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up vote 9 down vote accepted

Abel-Ruffini says that your polynomial has roots not expressible in terms of radicals. This is a much stronger condition than algebraicity: the field of algebraic numbers is a very large extension of the field of algebraic numbers expressible in radicals. This is in fact the very point of Abel's theorem.

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Incidentally, I wanted to say the algebraic numbers are of infinite degree over the radical-expressible numbers, but found I couldn't justify that. Anybody know? –  Kevin Carlson Aug 28 '12 at 15:36
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It's enough to show that for every prime $p$ there's an irreducible polynomial in $\mathbb{Q}[x]$ with Galois group $S_p$, which I'm reasonably sure is true. –  Chris Eagle Aug 28 '12 at 16:11
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There isn't a contradiction here, but many people do get confused on this point. It is true that the Fundamental Theorem of Algebra allows us to factorize any polynomial into a product of quadratics; but here is the catch: the Fundamental Theorem of Algebra doesn't guarantee that the coefficients of the quadratics can be expressed in terms of radicals (remember that the Abel-Rufini Theorem states that there exist polynomial equations whose solutions cannot be expressed exactly with radicals.) All this assumes that we are working over $\mathbb{R}$.

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The roots of $x^5-x+1=0$ are by definition algebraic. It is the product of a degree $1$ real polynomial, and two real quadratics. If we choose these factors to be monic (lead coefficient $1$), as we always can, then the coefficients of these three polynomials are algebraic.

This generalizes. Let $P(x)$ be a polynomial with real algebraic coefficients. Then in any factorization of $P(x)$ as a product of monic polynomials, all the coefficients of these polynomials are algebraic.

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